Question

While solving the equation \(e^{-x} - x = 0\) for \(x\), using the Newton-Raphson method with initial guess \(x_0 = 2\), the solution after the first iteration (\(x_1\)) is ________ (Rounded off to two decimal places)

Hint:

Apply the Newton-Raphson update \(x_1 = x_0 - f(x_0)/f\prime(x_0)\) with \(f(x) = e^{-x} - x\).

Answer 1

Correct Answer: 0.36

Step 1: Find the slope function first.
$f(x) = e^{-x} - x$ needs its derivative before anything else: $f'(x) = -e^{-x} - 1$, which is always negative.

Step 2: Evaluate the slope at $x_0 = 2$.
$e^{-2} \approx 0.135335$, so $f'(2) = -0.135335 - 1 = -1.135335$.

Step 3: Evaluate the function value at $x_0 = 2$.
$f(2) = 0.135335 - 2 = -1.864665$.

Step 4: Apply one Newton-Raphson step.
$x_1 = x_0 - f(x_0)/f'(x_0) = 2 - (-1.864665)/(-1.135335)$.
The correction term is $1.864665/1.135335 = 1.6424$, a positive number since the signs cancel.
So $x_1 = 2 - 1.6424 = 0.3576$.

Final Answer:
After one iteration, $x_1$ rounds to \[ \boxed{0.36} \]